5
$\begingroup$

By calculating the circulation per area of a vector field

$$F(x,y,z) = F_x(x,y,z)\vec{x} + F_y(x,y,z)\vec{y} + F_z(x,y,z)\vec{z}$$

in a small rectangle around $(x_0, y_0, z_0)$ on the $xy$ plane, it can be shown the limit as the sides of the rectangle approach zero is

$$\left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y}\right)$$

The same calculation however is not that straightforward if the rectangle does not lie in the $xy$, $yz$, or $xz$ planes. Now if $\vec{n}$ is the normal of the plane, I thought that by performing a change of basis such that $\vec{n} \rightarrow \vec{z'} $ and by following the previous calculations we could show that the limit of the circulation per area is

$$ \left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x_0, y_0, z_0)}{\partial y'}\right) $$

This is also the inner product of the curl of the vector field and the normal $\vec{n}$

As such the two should be equal:

$$\left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x'_0, y'_0, z'_0)}{\partial y'}\right) = \\ \left[\left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial y} - \frac{\partial F_y(x_0, y_0, z_0)}{\partial z} \right)\vec{x} + \left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial z} \right)\vec{y} + \left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y} \right)\vec{z}\right] \cdot \vec{n} $$ I've been trying to prove the above equality for some time without success, specifically I am not sure how to handle the transformations correctly. Any help with this is much appreciated!

$\endgroup$
4
+150
$\begingroup$

A simpler approach is via integral theorems. As stated in the question, the special cases for a rectangle in the $xy$ , $yz$ , $zx$ planes are well understood. According to Green's theorem : $$ \begin{cases} \iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy = \oint_{xy} \left( F_x\, dx + F_y\, dy \right) \\ \iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz = \oint_{xy} \left( F_y\, dy + F_z\, dz \right) \\ \iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx = \oint_{xy} \left( F_z\, dz + F_x\, dx \right) \end{cases} $$ But instead of rectangles, we take half rectangles, or better: the triangles $OAB$ , $OBC$ , $OAC$ respectively:

enter image description here

Thanks to Green's theorem we can replace area integrals by line-integrals; mind that they are counter-clockwise. Then it is clear that, irrespective of any further content: $$ \oint_{OAB} + \oint_{OBC} + \oint_{OAC} + \oint_{ABC} = 0 $$ Assuming that the operator rot(ation) is not defined yet in general, this means that we now have an expression for it: $$ 2 \iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA = \\ - \iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy - \iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz - \iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx $$ Continuing with infinitesimal volumes / areas and flipping normals on the right hand side, so that they become the components of the normal at the left hand side: $$ \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, \Delta A = \\ \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right)\cdot n_x\, \Delta A + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\cdot n_y\, \Delta A + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\cdot n_z\, \Delta A $$ Leaving out the infinitesimal area $\,\Delta A\,$ gives us the same answer as found by the OP themselves.
A somewhat neater approach is to calculate mean values and let the area of the (red) triangle go to zero: $$ \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n} = \lim_{ABC \to 0} \frac{\iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA}{\iint_{ABC} dA} $$ Note. I've encountered essentially the same method at several places elsewhere in physics (I think it's with stress and strain). Aanyway, a related subject is : What does shear mean?

$\endgroup$
  • $\begingroup$ Great answer! This is the same approach taken here file.scirp.org/pdf/APM20120100008_94595561.pdf no? I decided to go with rectangle loops because it is possible to rigorously prove Stroke's theorem using Riemann sums if we prove the density limit for rectangle loops. The problem with triangles for me is that there is a fundamental restriction on the triangle forms for each plane and it's harder to prove that the surface can be appropriately triangulated. $\endgroup$ – Veritas Sep 29 '16 at 21:37
  • $\begingroup$ Yes, I think that publication reflects some essentials of the method. But what physicists usually do is: simply delete the volume integral signs by an "infinitesimal" argument, as is shown in my last step. Being a physicist by education myself, I find this sufficient rigor already; mathematicians may have a different opinion about it. Though it employs triangles, yet I don't think my argument has anything to do with "triangulation". $\endgroup$ – Han de Bruijn Sep 30 '16 at 13:46
1
$\begingroup$

We start by applying a rotation around the x and y axis

$$ \left(\begin{array}{c} x' \\ y' \\ z' \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \left(\begin{array}{c} x \\ y \\ z\end{array}\right) $$

This rotates the surface so that its normal at the required point, points upwards. This means,

$$\left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \vec{n} $$

and by using the inversion property of rotation matrices,

$$ \vec{n} = \left(\begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i\end{array}\right) \cdot \left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} g \\ h \\ i \end{array}\right) $$

Notice that the normal $\vec{n}$ is the last row of our rotation matrix. Since the first and second row are also unit vectors orthogonal to $\vec{n}$,

$$ n = \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ d & e & f \end{array}\right| \\ n_x = \left|\begin{array}{cc}b & c \\ e & f \end{array}\right|, n_y = \left|\begin{array}{cc}d & f \\ a & c \end{array}\right|, n_z = \left|\begin{array}{cc}a & b \\ d & e \end{array}\right|$$ We also want to rotate our vector field appropriately: $$ \left(\begin{array}{c} F_{x'}(P') \\ F_{y'}(P') \\ F_{z'}(P') \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \left(\begin{array}{c} F_x(P) \\ F_y(P) \\ F_z(P)\end{array}\right) $$

Finally, $$\begin{gather}\frac{\partial F_{y'}}{\partial x'} - \frac{\partial F_{x'}}{\partial y'} \end{gather} = \\ \left|\begin{array}{cc} \frac{\partial}{\partial x'} & \frac{\partial}{\partial y'} \\ F_{x'} & F_{y'} \end{array}\right| = \\ \left|\begin{array}{ccc} a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y} + c\frac{\partial}{\partial z} & d\frac{\partial}{\partial x} + e\frac{\partial}{\partial y} + f\frac{\partial}{\partial z} \\ aF_{x} + bF_{y} + cF_{z} & dF_{x} + eF_{y} + fF_{z} \end{array}\right| = \\ \left| \begin{array}{cc} a\frac{\partial}{\partial x} & e\frac{\partial}{\partial y} \\ aF_x & eF_y\end{array}\right| + \left| \begin{array}{cc} a\frac{\partial}{\partial x} & f\frac{\partial}{\partial z} \\ aF_x & fF_z\end{array}\right| + \left| \begin{array}{cc} b\frac{\partial}{\partial y} & d\frac{\partial}{\partial x} \\ bF_y & dF_x\end{array}\right| + \\ \left| \begin{array}{cc} b\frac{\partial}{\partial y} & f\frac{\partial}{\partial z} \\ bF_y & fF_z\end{array}\right| + \left| \begin{array}{cc} c\frac{\partial}{\partial z} & d\frac{\partial}{\partial x} \\ cF_z & dF_x\end{array}\right| + \left| \begin{array}{cc} c\frac{\partial}{\partial z} & e\frac{\partial}{\partial y} \\ cF_z & eF_y\end{array}\right| = \\ (bf-ce)\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| + (af-cd)\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + (ae-db)\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\ \left|\begin{array}{cc}b & c \\ e & f \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| - \left|\begin{array}{cc}d & f \\ a & c \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + \left|\begin{array}{cc}a & b \\ d & e \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\ n_x\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| - n_y\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + n_z\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\ \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{array}\right| \cdot \vec{n}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.